If I use that $\displaystyle \Gamma \left( \displaystyle \frac{1}{ n}\right) \approx n$ when $n$ is large, then I wonder if it's possible to compute the following limit in a closed-form

$$\lim_{n\to \infty}\left(\frac{1}{ \Gamma\left(\displaystyle \frac{1}{1}\right)}+ \frac{1}{ \Gamma\left( \displaystyle \frac{1}{2}\right)}+ \cdots + \frac{ 1}{ \Gamma \left( \displaystyle \frac{1}{ n}\right) }- \log\left( \Gamma\left(\displaystyle\frac{1}{n}\right)\right)\right),$$
where I called $\displaystyle \sum_{k=1}^{\infty}\frac{ 1}{ \Gamma \left( \displaystyle \frac{1}{ k}\right) }$ as Gammaharmonic series.
I can get approximations, but I cannot get the precise limit, and I don't even know if it can be expressed in terms of known constants.

A 500 points bounty moment: I would enjoy pretty much finding a solution (containing a closed-form) for the posed limit, hence the generous bounty. It's unanswered for 3 years and 8 months, and it definitely deserves another chance. Good luck!

$\begingroup$This is a very interesting question, indeed ! Numerically, it seems that there is a limit but how to express it, that is the question !$\endgroup$
– Claude LeiboviciSep 27 '14 at 16:53

1

$\begingroup$Wrench (1968) has a paper concerning the series $\sum\frac{1}{\Gamma(k)}$. Some answers are already showing some promise to computing that series.$\endgroup$
– Ali CaglayanSep 27 '14 at 22:56

$\begingroup$Are you sure that your question can be answered during 7 days although it's unanswered for 3 years and 8 months ? I think the answers of Leucippus and Jack D'Aurizio are well done! :-)$\endgroup$
– user90369May 29 '18 at 11:56

$\begingroup$@user90369 no problem then. In general, one user will remain with my legacy, that is the bounty, and I'm sure they (those receiving such bounties) will want to continue to work on my question, maybe for months, years (if the case, of course), until they finally get the solution (at least out of curiosity). :-)$\endgroup$
– user 1357113May 29 '18 at 13:04

From the Weierstrass product for the Gamma function we have, as $x\to+\infty$:
$$\frac{1}{\Gamma(1/x)}=\frac{1}{x}+\frac{\gamma}{x^2}+O\left(\frac{1}{x^3}\right)\tag{1}$$
and:
$$\log\Gamma(1/x)=\log x -\frac{\gamma}{x}+O\left(\frac{1}{x^2}\right)\tag{2}$$
gives that the value of the limit is:
$$\gamma+\sum_{n=1}^{+\infty}\left(\frac{1}{\Gamma(1/n)}-\frac{1}{n}\right)=0.8188638872713\ldots\tag{3}$$

Lemma 1.
If $f(x)$ is analytic in $(-a,a)$, $a\geq 1$, then
$$
\sum^{M}_{n=1}f\left(\frac{1}{n}\right)=\int^{M}_{1}f\left(\frac{1}{t}\right)dt+c(f)+O\left(\frac{1}{M}\right)\textrm{, }M\rightarrow +\infty,\tag 1
$$
where $c_f$ is a constant depended from $f$ and not from $M$:
$$
c_f=f(0)+f'(0)\gamma+\sum^{\infty}_{k=2}\frac{f^{(k)}(0)}{k!}\left(\zeta(k)-\frac{1}{k-1}\right).
$$
Proof.

Expand $f$ in (1) into Taylor series, then sum and integrate. The ''infinite'' terms involving $M$ are canceled and the result will follow.

Remark. The next two known estimates are usefull in the proof of Lemma 1:
$$
\sum^{M}_{k=1}\frac{1}{k}=\log(M)+\gamma+O\left(\frac{1}{M}\right)\textrm{, }M\rightarrow \infty
$$
and
$$
\zeta_n(s)-\zeta(s)=O\left(\frac{1}{n^{s-1}}\right)\textrm{, }s>1\textrm{, }n\rightarrow\infty.
$$